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## Karnataka 2nd PUC Statistics Model Question Paper 1 with Answers

Time: 3 Hours 15 Minutes

Max. Marks: 100

Section – A

I. Answer any Ten of the following questions: (10 × 1=10)

Question 1.

What is Longevity in a life table?

Answer:

Life expentancy of a new – bom baby is called longevity.

Question 2.

Which index number shows upward bias?

Answer:

Laspeyre’s.

Question 3.

Name the index number which satisfies both Time Reversal Test and Factor Reversal Test.

Answer:

Fisher’s Index number

Question 4.

What is Historigram?

Answer:

Graph of actual values of time series is called Historigram.

Question 5.

Define a Bernoulli variate.

Answer:

p(x) = p^{x} q^{1-x}; x = 0,1, 0 < p < 1; Here ‘x’ is called Bernoulli variate.

Question 6.

If n = 10 for t – distribution, find its variance.

Answer:

Question 7.

Write down the standard error of sample proportion p.

Answer:

Question 8.

Which is the estimator of the population mean?

Answer:

x̄

Question 9.

Define statistical hypothesis.

Answer:

A statements / assertion made regarding the parameter is called statistical hypothesis.

Question 10.

Which type of variation can be detected with the help of statistical quality control?

Answer:

Variation due to assignable causes.

Question 11.

What do you mean by value of a game?

Answer:

The pay off at the saddle point is called the value of a game.

Question 12.

Mention one advantage of Inventory.

Answer:

If helps in smooth and efficient running of the bussines

Section – B

II. Answer any Ten of the following questions: (10 × 2 = 20)

Question 13.

Mention the methods of collection of vital statistics.

Answer:

- Census method and
- Registration method

Question 14.

State two limitations of index numbers.

Answer:

- Many formulae are used which gives different answers for the index.
- Index numbers are based on sample data, so, they are only approximate indicators.

Question 15.

If Laspeyre’s price index number is 124 and Paasche’s price incex number is 126, find Fisher’s price index number.

Answer:

Question 16.

State the different phases of ‘Business Cycle’.

Answer:

- Prosperity
- Decline
- Depression
- Recovery / Improvement.

Question 17.

Write down the two assumptions of interpolation and extrapolation.

Answer:

- There are no sudden Jumps in the dependent variable from one period to another.
- There will be no consecutive missing values in the series.

Question 18.

If a = 7, b = 9 and n = 5, find the variance of hyper- geometric distribution.

Answer:

Question 19.

For a chi – square variate (χ^{2}) with 12 degrees of freedom P[0 < χ^{2} < 11.34] = 0.5. Find median and mode.

Answer:

Given that p(0 < χ^{2} < 11.34) = 0.5

∴ Median = 11.34, Given n = 12 degrees of freedom

So, mode = n – 2 = 12 – 2 = 10.

Question 20.

Define sample space and parameter space.

Answer:

Sample space : It is the set of all samples of same size drawn from the population.

Parameter space : It is the set of all admissible values of parameters.

Question 21.

What do you mean by

a) Critical region b) Power of a test?

Answer:

‘Critical region’ is the set of all those values of test statistic which lead£4o the rejection of the null hypothesis.

‘Power of a test’ is the probability of rejection null hypothesis when it is not true. (1 – β)

Question 22.

Mention two disadvantages of acceptance sampling plan.

Answer:

- There is a risk of accepting bad lot and rejecting a good lot, since verification is made only on the basis of samples.
- Timely identification of the production of defectives cannot be achieved.

Question 23.

In a ‘two – person zero – sum game’ the pay – off matrix of player A is given below.

Write down the pay – off matrix of player B.

Answer:

Question 24.

State the assumption of E.O.Q model – II.

Answer:

- There is a uniform demond
- Production is instantaneous
- Shortages are not allowed
- Lead times zero

Section – C

III. Answer any Eight of the following questions: (8 × 5 = 40)

Question 25.

Calculate gross reproduction rate for the following data.

Answer:

W.S.F.R (15 – 19 ) = \(\frac{200}{1000}\) × 1000 = 20 ; Similarly for other age groups W.S.F.Rs can be calculate.

= 5 ΣWSER = 5 × 270.36 = 1351.8 Female births 1000 women.

Question 26.

Define index number. Mention four uses of index numbers.

Answer:

Index number is a statistical measure designed to show an average change in a variable or group of related’variables over a period (geographical area / Income group)

- Index numbers are used by Government in formulating Economic / Business policies, such as Taxation, imports / Exports, Fixation of bank rates etc.
- Infixation of salary and grant of dearness allowances to employees.
- In evaluation of real rupee value.
- Comparing varaiation in prices / Quantity / Demand / supply of goods.

Question 27.

Calculate cost of living index number for 2010 with base 2005.

Answer:

Question 28.

Compute 5 yearly moving averages for the data given below.

Answer:

Let x and y be the year and value.

Question 29.

In the following table the values of X represents the degrees of freedoip and the Y values represents χ^{2} – critical values at 5% level of significance. Find the missing values by using binomial expansion method.

Answer:

The number of known values of y ‘n’ = 5; The Binomial exponsion (y – 1)^{5} = 0. So,

(y – 1)^{5} = y_{5} – 5y_{4} + 10y_{3} – 10y_{2} + 5y_{1}+ y_{0} = 0 → (1)

14.07-5 (12.59) + 10y_{3} – 10 (9.49) + 5 (7.81) – 5.99 = 0

14.07 – 62.95 + 10y_{3} – 94.9 + 39.05 – 5.99 = 0

10y_{3} – 110.77 = 0; ∴ y_{3} = 11.077

To get y_{6} the suffixes of ‘y’ has increased by ‘1’ in equation (1); we get

y_{6} – 5y_{5} + 10y_{4} – 10y_{3} + 5y_{2} – y_{1} = 0

y_{6} – 5 (14.07) + 10(12.59) – 10 (11.077) + 5 (9.49) – 7.81 = 0

y_{6} – 70. 35 + 125.9 – 110.77 + 47.45 – 7.81 = 0 ; y_{6} – 15.58, ∴ y_{6} = 15.58

Question 30.

In a college, 70% of the students are boys. In a random sample of 5 students find the probability of getting

a) 4 boys b) at the most 2 boys.

Answer:

Let x be the number of boys is a Binomial variate with the parameters n = 5; p = 70% = 0.7; q = 1 – p = 0.3

Then the p.m.f is; p(x) = ^{n}c_{x} p^{x} q^{n-x} ; x = 0,1,2 ……. n

p(x) = ^{5}c_{x} 0.7^{x} 0.3^{5-x}; x = 0,1,2…5

(i) p (4 are boys) = ^{5}c_{4}(0.7)^{4}(0.3)^{5-4} = 5 × 0.2401 × 0.3 = 0.3602

(ii) p (at the most 2 boys) = p(x ≤ 2) = p(x = 0) + p(x = 1) + p(x = 2)

= ^{5}c_{0} 0.7^{0} 0.3^{5-0} + ^{5}c_{1} 0.7^{5-1} 0.3^{5-1} + ^{5}c_{2} 0.7^{2} 0.3^{5-2}

= (1 × 1 × 0.0024) + (5 × 0.7 × 0.081) + (10 × 0.343 × 0.027)

= 0.0024 + 0.0284 + 0.1323 = 0.1631

Question 31.

Write down 5 properties of normal distribution.

Answer:

- Curve is bell shaped : Mean = Median = mode.
- The curve’is symmetric ie., Non – skew (β
_{1}= 0) - The curve is mesokurtic (β
_{2}= 3) - Mean = µ variance = σ
^{2}; SD = σ - Q.D = \(\frac { 2 }{ 3 }\) σ ; MD = \(\frac { 4 }{ 5 }\) σ
- Area under the Normal curve = 1

Question 32.

A sample of 50 children is taken from a school. The average weight of the children is 28 kg and S.D is 5kg. Test at 1% level of significance, whether the average weight of the school children is less than 30kg.

Answer:

Given : n = 50, x̄ = 28, s = 5, α = 1% µ = 30

H_{0 }: Average weight of school children is 30kg (ie, µ = 30kg)

H_{1} : Average weight of school children is less than 30 kg (ie, µ < 3kg) {Lower tail test}

At α = 1 % the lower tail critical value -k = -2.33

Here z_{cal} < – k / Z_{cal} lies in Rejection Region (R.R)

∴ H_{0} is rejected and H_{1} is accepted. .

Conclusion : Average weight of the school children is less than 30 kg (µ < 30)

Question 33.

From the following data, test whether literacy and smoking are independent at 5% level of significance.

Answer:

H_{0}: Literacy and smoking are independent.

H_{1}: Literacy and smoking are dependent {upper tail test}

The given data can be written as in a 2 × 2 contigency table as below

At α = 5% for 1.d.f. the upper tail critical value K_{2} = 6.63

Here χ^{2}_{cal} < k_{2} / χ^{2}_{cal} lies in Acceptance region (A.R). ∴ H_{0} is accepted.

Conclusion: Smoking and literacy are independent.

Question 34.

Following table gives mean (X̄) and range (R) or 6 sample of size 4 each.

Find the control limits for drawing X̄ – chart.

Answer:

The control limits for x̄ chart are: C.L = x̿ = 13.1833

L.C.L = x̿ – A_{2}R̄ = 13. 1833 – (0.729 × 7) = 8.0803

U.C.L = x̿ + A_{2}R̄ = 13.1833 + (0.729 × 7) = 18.2863

Question 35.

Solve the following L.P.P graphically :

Minimum Z = 5x + 8y

Subject to 3x + 2y ≤ 18

4x + 3y ≥ 12

and x ≥ 0, y ≥ 0

Answer:

Consider the constraints as equabties and obtain two co – ordinates and plot the points on xy – plane to get straight lines.

consider ; 3x + 2y= 18; put x = 0 ; y = 9; (0,9) as co – ordinates.

put y = 0; x = 6; (6,0)

4x + 3y = 12;

Non – negative restriction : put x = 0, y = 4; (0,4)

put y = 0,x = 3;(3,0) and

x = 0, y = 0

From the graph the feasible region exists at the comer points A(0,4), B (0,9), C (6,0), D (3,0)

The optimum value of the objective comer points are obtained as below:

The objectivity of L.P.P Minimises at the comer point D(3,0). The suggested optimum Solution is are: x = 3, y = 0. The optimum value Z_{D} = 15.

Question 36.

Find an initial basic feasible solution by matrix minima method for the following transportation problem and also find the total cost associated with the basic feasible solution.

Answer:

Lrt a_{1} and b_{1} be supply and demand

By Matrix minima method first allocation is made at the least cost 14 at (3,2) as;

X_{12} = Min (24, 12) 12 ‘C’ is satisfied and so delete. Replace 24 by (24 – 12) = 12. Write down the reduced cost matrix.

The least cost is 18, allocate as : X_{22} = Min (48,44) = 44, B is satisfied and so delete. Replace 48 by (48 – 44) = 4. Write down the reduced cost matrix

The least is 19, alLocate as: X_{11} = Mm (40,56) = 40 ‘I’ is satisfied and so delete. Replace 56 by (56 – 40) = 1.

Remainîng cost matrix is :-

Allocate at the least cost as : X_{31} = Min (12,16) = 12, III is ignored and (16 – 12) = 4, the remaining allocation is made as; x_{21} = Min (4,4) = 4

The suggested I.B.F.S and the total cost is :

Section – D

IV. Answer any Two of the following questions: (2 × 10 = 20)

Question 37.

Compute standardised death rates for village A and village B. State which village is healthier.

Answer:

Let A and B be the ASDRs of villages A and B and ‘p be the standard population

For village A: ASDR (0 – 20) = \(\frac{128}{8000}\) × 1000 = 16

For village B: ASDR (0 – 20) = \(\frac{72}{4000}\) × 1000 18;

similarily ASDRs can be calculated for other age groups

SDR (A) = \(\frac{\Sigma P_{S} A}{\Sigma P_{s}}=\frac{1905000}{100,000}\) = 19.05 Deaths/ 1000 population

SDR(B) = \(\frac{\Sigma P_{\mathrm{S}} B}{\Sigma P_{\mathrm{s}}}=\frac{1675000}{100,000}\) = 16.75 Deaths! 1000 population.

Here SDR (B) < SDR(A) and so, village B ¡s healthier than village A.

Question 38.

Construct Marshall – Edgeworth, Fisher’s and Dorbish – Bowley’s price index numbers

Answer:

Question 39.

Fit an exponential trend of the form y = ab^{x}. Also estimate the value for 2008.

Answer:

Let x and y be the year and value, here x̄ = 2003

From the Normal equation: nloga + logb. Σx = Σlog y

Since Σx = 0; log a = \(\frac{\Sigma \log \mathrm{y}}{\mathrm{n}}=\frac{7.853}{7}\) = 1.1219

∴ a = Anti log (log a); a = A.L (1.1219) = 13.2404

And from: log Σ x + log Σb Σx^{2} = Σx logy

log b = \(\frac{\Sigma x \cdot \log y}{\Sigma x^{2}}=\frac{4.2068}{28}\) = 0.1502

∴ b = Antilog(logb) = A.L (0.1502) = 1.4132

The fitted exponential trend equation is :- y = ab^{x}; ŷ = 13.2404 (1.44 32)x

Estimation for the year 2008: x = 5; ŷ(2008) = 13.2404(1.4132)^{5} = 74.631 units.

Question 40.

Fit a Poisson distribution to the following data and test for goodness of fit at 5% level of significance.

Answer:

Given : Let x be the number of mistakes is a poisson variate, the parameter X is obtained as below.

Let x be the number of pages, then from the distribution : Mean = x̄ = λ = \(\frac{\Sigma \mathrm{fx}}{\mathrm{N}}\)

Using Recurrence the successive theoretical frequencies can be obtained as below

∴ T_{5} and more = 100 – [ 30 + 36 + 22 + 9 + 3] = 100 – 100 = 0

The fitted observed and theoretical frequency distribution is:

Chi-square Test:

H_{0} : poisson distribution is good fit (ie., O_{i} = E_{i})

H_{1} : poisson distribution is not good fit (ie, O_{1} ≠ E_{1}) (upper tail test}.

Here d.f will be (n – 1 – 1) = n – 2 because λ is computed from the data.

χ^{2}_{cal} = 0.5231

At α = 5% for (n – 2) = (4 – 2) = 2d. f the upper tail critical value k_{2} = 5.99

Here χ^{2}_{cal} lies in Acceptance region ∴ H_{0} is accepted.

Conclusion : Poisson distribution is good fit.

Section – E

V. Answer any Two of the following questions: (2 × 5 = 10)

Question 41.

The daily wages of workers of a factory are normally distributed with mean Rs.700 and S.D. Rs. 40. Find the probability of workers whose daily wages will be

a) more than Rs. 800 b) between Rs. 690 and Rs.720

Answer:

Given : Let x be the daily wages is a Normal variate with Mean: p = 700, and SD :

σ = 40, then the S.N. V is :- Z = \(\frac{x-\mu}{\sigma}\) ~ N(0,1) ; Z = \(\frac{x-700}{40}\)

(i) P(x > 800) = p(z > \(\frac{800-700}{40}\)) = p(z > 2.5)

= Area from 2.5 to ∞ = 0.0062

(ii) p(690 < x < 720) = p\(\left(\frac{690-700}{40}<z<\frac{720-700}{40}\right)\)

= p(-0. 25< z < 0.5)

= Area from -0. 25 to ∞ – Area from 0.5 to ∞

= 0.5987 – 0.3085 = 0.2902

Question 42.

For the following data test whether there is any significant difference in the population proportions at 5% level of significance.

Answer:

Given: n_{1} = 100; P_{1} = 0.02, n_{2} = 110, p_{2} = 0.01

H_{0}: There is no significant difference in the population proportions (ie., p_{01} = p_{02})

H_{1}: There is a significant difference in the population proportions ie., p_{01} ≠ p_{02} {Two tail test}

At α = 5% the two tail critical values are ±k = ± 1.96

Here Z_{cal} lies in acceptance region

∴ H_{0} is accepted Conclusion : p_{01} = p_{02}

Question 43.

The marks scored by 9 students in two tests conducted before coaching and after coaching are as follows.

Whether coaching is effective.

Answer:

H_{0} : Marks before and after coaching is same (µ_{1} = µ_{2})

H_{1} : Marks after coaching has improved (µ_{1} < µ_{2}) {Lower tail-test}.

Let x and y be the marks before and after coaching

At α = 5% for (n – 1) = 9 – 1 = 8 d.f the lower tail critical value k = -1.86

Here t_{cal} lies in acceptance region (R.R)

∴ H_{0} is rejected H_{1} is accepted

Conclusion : µ_{1} > µ_{2}

Question 44.

A firm is considering of replacing certain equipment whose purchase cost is Rs.5000. Suggest a suitable period to replace the equipment.

Answer:

Given p = 5000

From the above table the annual average cost is minimum for the 4th year; is the optimum period for replacement of the equipment: A(4) = Rs. 2700